What is the relationship between force, acceleration, and mass in a lift?

In summary, the question involves finding the mass of a body that is hanging from a spring-balance in a lift. The balance reads 70N when the lift has an upward acceleration of 4.0ms^-2. After multiple attempts, the correct method is to divide the weight of 70N by the sum of the upward acceleration and gravitational field strength (14ms^-2). This results in a mass of 5kg. To explain why this works, one must use Newton's 2nd Law which states that the unbalanced force on an object is equal to its mass multiplied by its acceleration. The unbalanced force in this case is the difference between the tension in the spring and the weight of the body. By setting
  • #1
TH02
18
4
Homework Statement
A body hangs from a spring-balance which is suspended from the ceiling if a lift. What is the mass of the body if the balance registers a reading of 70N when the lift has an upward acceleration of 4.0ms^-2. (g = 10ms^-2)
Relevant Equations
F = ma
Weight = mg
Much like the previous question I've posted I was quite unsure of the the method I should take and resorted after many attempts to just trial and error which resulted in getting the correct answer of 5kg after dividing the weight of 70N by 14ms^-2 (sum of upward acceleration and gravitational field strength). Can someone please explain why this works, or the proper method to the question?
 
Physics news on Phys.org
  • #2
TH02 said:
Homework Statement:: A body hangs from a spring-balance which is suspended from the ceiling if a lift. What is the mass of the body if the balance registers a reading of 70N when the lift has an upward acceleration of 4.0ms^-2. (g = 10ms^-2)
Relevant Equations:: F = ma
Weight = mg

Much like the previous question I've posted I was quite unsure of the the method I should take and resorted after many attempts to just trial and error which resulted in getting the correct answer of 5kg after dividing the weight of 70N by 14ms^-2 (sum of upward acceleration and gravitational field strength). Can someone please explain why this works, or the proper method to the question?
Why do you think it doesn't work?

What is the ##F## in ##F = ma##? And don't just say "force"!
 
  • #3
Would it be the tension in the spring? Or the weight of the body?
 
  • #6
Chestermiller said:
Let’s see your fbd.
IMG_20200703_162730234.jpg


I am not sure if I've added all the forces I need
 
  • #7
TH02 said:
View attachment 265762

I am not sure if I've added all the forces I need
Gravity? Isn't ##T = 70N##?
 
  • #8
PeroK said:
Gravity? Isn't ##T = 70N##?
I'm sorry, I'm getting quite confused
 
  • #9
TH02 said:
I'm sorry, I'm getting quite confused
The question tells you that the spring balance reads ##70N##. That means that ##T = 70N## in your diagram.
 
  • #10
PeroK said:
The question tells you that the spring balance reads ##70N##. That means that ##T = 70N## in your diagram.
IMG_20200703_164919018.jpg

Is this more accurate?
 
  • #11
That looks better. What does ##a = 4m/s^2## tell you?
 
  • #12
PeroK said:
That looks better. What does ##a = 4m/s^2## tell you?
That I have a vertical acceleration of 4m/s^2?
 
  • #13
TH02 said:
That I have a vertical acceleration of 4m/s^2?
Yes, okay, but more than that. Hint: Newton's 2nd Law.
 
  • #14
PeroK said:
Yes, okay, but more than that. Hint: Newton's 2nd Law.
I'm not following, is it that the force being applied to the body is directly proportional to the acceleration acting on the body?
 
  • #15
TH02 said:
I'm not following, is it that the force being applied to the body is directly proportional to the acceleration acting on the body?

The ##F## in Newton's law is the unbalanced force, which is the vector sum of all forces on the body. In this case we have:
$$F = T - mg, \ \ \text{and} \ \ F = ma$$
Where, to be precise, I have taken upwards as the positive dircetion.
 
  • Like
Likes Lnewqban
  • #16
PeroK said:
The F in Newton's law is the unbalanced force, which is the vector sum of all forces on the body. In this case we have:
F=T−mg, and F=ma
Where, to be precise, I have taken upwards as the positive dircetion.
Ahhh I see
IMG_20200703_171831782.jpg

Sorry for taking so long to get there! I appreciate the help
 
  • Like
Likes Lnewqban and PeroK
  • #17
TH02 said:
I'm not following, is it that the force being applied to the body is directly proportional to the acceleration acting on the body?
 
Back
Top